Some Wolfram|Alpha Features I can't achieve in Mathematica

Question

I have noticed some features available on the site Wolfram|Alpha which I am not managing to replicate. Two examples which come to mind are the following.

1. Given a function $f(x)$, I want to determine its Taylor series, and its radius of convergence also. For example, take $f(x)=\ln(1-x^2)$. On Wolfram|Alpha, I get the following output:

   

   where I get the nice "converges when $|x|<1$" detail below. I know that I can obtain the series by running the command Series[Log[1-x^2], {x,0,8}], but I don't know how to get the radius of convergence. The inbuilt command SumConvergence does not allow one to determine the radius, since the general term of the series must be known.

2. Suppose we have a homogeneous system of linear equations represented by the matrix equation $\mathbf{Ax}=\mathbf0$, e.g. $$\underbrace{\begin{pmatrix} 3&2&1\\7&6&k\\k&4&3 \end{pmatrix}}_{\mathbf A}\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix} 0\\0\\0\end{pmatrix}$$ Both the Solve and LinearSolve commands give only the trivial solution $x=y=z=0$ as the solution to the system. But the website gives the interesting ones too:

   

There are a few other situations where I have found the site more useful than the application - but so far these two have been the most frustrating. Is there any way to achieve the two above in Mathematica without having to go online?

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EDIT For 2, I attempted to use Reduce by running the following code.

Reduce[{
 {3, 2, 1},
 {7, 6, k},
 {k, 4, 3} } . {x,y,z} == {0, 0, 0}]]

but got the following (better, but still not as good) output:

(z == 0 && y == 0 && x == 0) || ((y == -2 z || y == -((5 z)/4)) && x == 1/3 (-2 y - z) && y != 0 && k == (10 (2 y + z))/(3 y))

Answer 1

To shed some light why Mathematica only gives the trivial NullSpace for example 2 in post, I solved this by hand and this is what I found:

$$ A= \begin{pmatrix} 3 & 2 & 1\\ 7 & 6 & k\\ k & 4 & 3 \end{pmatrix} $$

To find the null space, we solve for $x$ in $Ax=0$. Applying Gaussian elimination

$$ \begin{pmatrix} 3 & 2 & 1\\ 7 & 6 & k\\ k & 4 & 3 \end{pmatrix} \overset{R_{2}=R_{2}-\frac{7}{3}R_{1}}{\underset{R_{3}=R_{3}-\frac{k}{3} R_{1}}{\longrightarrow}} \begin{pmatrix} 3 & 2 & 1\\ 0 & \frac{4}{3} & k-\frac{7}{3}\\ 0 & 4-\frac{2}{3}k & 3-\frac{k}{3} \end{pmatrix} $$

Now $R_{3}=R_{3}-\frac{\left( 4-\frac{2}{3}k\right) }{\frac{4}{3}}R_{2}$ gives

\begin{equation} \begin{pmatrix} 3 & 2 & 1\\ 0 & \frac{4}{3} & k-\frac{7}{3}\\ 0 & 0 & \frac{1}{2}k^{2}-\frac{9}{2}k+10 \end{pmatrix} \tag{1} \end{equation}

Last row gives

$$ \left( \frac{1}{2}k^{2}-\frac{9}{2}k+10\right) z=0 $$

Two cases to consider in (1)

case 1 If $\frac{1}{2}k^{2}-\frac{9}{2}k+10\neq0$ then $z=0$. Second row now gives $\frac{4}{3}y+\left( k-\frac{7}{3}\right) z=0$, or $y=0$. First row gives $x=0$. Hence in this case the NullSpace is

$$ \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} $$

case 2 If $\frac{1}{2}k^{2}-\frac{9}{2}k+10=0$, then $z=t$, a free parameter, which can be any value. Second row in (1) now gives \begin{align*} \frac{4}{3}y+\left( k-\frac{7}{3}\right) t & =0\\ y & =-\frac{3}{4}t\left( k-\frac{7}{3}\right) \end{align*} And first row gives

\begin{align*} 3x+2y+z & =0\\ 3x & =-2\left( -\frac{3}{4}t\left( k-\frac{7}{3}\right) \right) -t\\ x & =\frac{1}{2}t\left( k-3\right) \end{align*}

Hence in this case, the NullSpace is

$$ \begin{pmatrix} x\\ y\\ z \end{pmatrix} =t \begin{pmatrix} \frac{1}{2}\left( k-3\right) \\ -\frac{3}{4}\left( k-\frac{7}{3}\right) \\ 1 \end{pmatrix} $$

Since $\frac{1}{2}k^{2}-\frac{9}{2}k+10=0$, the solution is $k=5,k=4$. Substituting these values into the above gives the NullSpace as

$$ \begin{pmatrix} x\\ y\\ z \end{pmatrix} = t\left\{ \begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix} , \begin{pmatrix} \frac{1}{2}\\ -\frac{5}{4}\\ 1 \end{pmatrix} \right\} $$

For any $t$. For only $t=0$, this gives the result gives by Mathematica.

So Wolfram Mathematica only considered the case 1 above, and did not consider case 2. Wolfram Alpha, which has additional A.I. engine, seems to have looked more into this.